Question 1201232: A soft drink manufacturer claims that its 120-milliliter bottles do not contain, on average, more than 32 calories. A random sample of 20 bottles of this soft drink, which were checked for calories, contained a mean of 33.3 calories with a standard deviation of 2.5 calories. Assume that the number of calories in the 120-milliliter bottles is normally distributed. Using a level of significance of 5%, is there a significant difference between the manufacturer’s claim and the findings of the study? t computed = 2.32
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! population mean is assumed to be 32 calories
sample size is 20
sample mean is 33.3
sample standard deviation is 2.5
standard error is standard deviation / square root of sample size = 2.5 / sqrt(20) = .5590169944.
test t-score is (x - m) / s = (33.3 - 32) / .5590169944 = 2.325510697
area to the right of that at 19 degrees of freedom is equal to .0156354548.
x is the sample mean
m is the assumed population mean
s is the standard error.
at .05 two tailed level of significance, this is less than .025, making the results of the test significant.
if you use critical t-score, that would be equal to 2.093024022.
since that is less than the test t-score of 2.32....., the results are, once again, significant, as they should be, because the critical t-score and the critical level of significance always give results that are consistent with each other.
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